This is an exercise from Conway that I am stuck at. For (a) I can't understand the hint? How does the inequality in the hint holds? And what does it mean by the "maximum value?"
For (b) I think I have to use the definition of limit infimum, but can't proceed... Could anyone please help me?
(a) "maximum value" should be "minimum value".
If the order of $f$ is $\lambda $ and $\beta >\lambda $, $$ |f(z)|<\exp(|z|^\beta )$$ holds for all $z$ with $|z|$ sufficiently large. By Cauchy's estimate \begin{align} |c_n|&\le \frac{M(r)}{r^n} \quad (M(r)=\max_{|z|=r}|f(z)|)\\ &\le \frac{1}{r^n}\exp(r^\beta ). \end{align} Taking $r=(n/\beta )^{1/\beta }$ we have \begin{align} |c_n|\le \left(\frac{n}{\beta }\right)^{-\frac{n}{\beta}}\exp\left(\frac{n}{\beta }\right).\tag{1} \end{align} for sufficiently large $n$. Note that $\left(\frac{n}{\beta }\right)^{-\frac{n}{\beta}}\exp\left(\frac{n}{\beta }\right)$ is the minimum value of $r^{-n}\exp(r^\beta )$. From $(1)$ we have $$ \frac{-\log |c_n|}{n\log n}\ge \frac{1}{\beta }\left(1-\frac{1+\log \beta }{\log n}\right)\ge \frac{1}{2\beta } $$ for sufficiently large $n$. This implies $\alpha >0.$
(b) If $\alpha =\liminf_{n\to \infty} \frac{-\log |c_n|}{n\log n}$, by the definition of $\lim \inf$ there is an integer $p$ such that $$ \alpha -\varepsilon < \frac{-\log |c_n|}{n\log n}$$ for $n>p$. This leads to \begin{align} -(\alpha -\varepsilon )\log n&> \frac{1}{n}\log |c_n|,\\ n^{-(\alpha -\varepsilon )}&>|c_n|^{\frac{1}{n}}. \end{align} Hence we have \begin{align} |f(z)|&\le \sum_{n=0}^\infty |c_n|r^n=\sum_{n=0}^p |c_n|r^n+\sum_{n=p+1}^\infty |c_n|r^n\\ &\le \left(\sum_{n=0}^p |c_n|\right)r^p+\sum_{n=p+1}^\infty \left(\frac{r}{n^{\alpha -\varepsilon }}\right)^n. \end{align}